NOTES AND DISCUSSIONS
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More about the falling raindrop
Carl E. Mungana兲
Department of Physics, U.S. Naval Academy, Annapolis, Maryland 21402-1363
共Received 13 May 2010; accepted 7 July 2010兲
A simple strategy is presented for solving the “inverse rocket” problem of a particle accumulating
material from a medium through which it falls vertically. Some forms of drag can also be easily
included, thereby changing the constant acceleration to a more realistic value.
关DOI: 10.1119/1.3471338兴
Sokal1 has introduced a general version of the problem of
a raindrop accumulating mass as it falls through mist and
solved it using the chain rule and an integrating factor. In this
Note, it is shown that his solution can be simplified by using
momentum as the dependent variable, making the solution
more accessible to introductory physics majors. In addition,
the problem is further generalized to include a drag force.
In the absence of drag, Newton’s second law for the drop
is
dp
= mg,
dt
cases ␥ = 1 and ␤ = 0 共linear drag with speed-independent
mass accretion兲 and ␥ = 2 and ␤ = 1 共quadratic drag with linear speed accretion兲. The solution when ␥ = 1 + ␤ can be obtained by dividing Eq. 共5兲 by Eq. 共2兲, which leads to
dp ␧ p g 1+␤−␣ −␤
= m
+
p .
dm ␭ m ␭
The left-hand side of Eq. 共6兲 suggests changing the dependent variable to u = m␧/␭ p, which yields3
共1兲
where p = m␷ is its momentum and the gravitational field g is
assumed to be constant. The mass accretion rate is assumed
to scale with powers of the mass 共or size兲 and speed of the
drop,
dm
= ␭m␣␷␤ = ␭m␣−␤ p␤ ,
dt
g
p␤dp = m1+␤−␣dm,
␭
g
u␤du = m共1+␤兲共1+␧/␭兲−␣dm
␭
共3兲
␷1+␤ =
2+␤−␣
p
C
gm
+
=
,
1 + ␤ ␭共2 + ␤ − ␣兲 1 + ␤
a=
共4兲
where the last term is a constant of integration. If we substitute p = m␷, Eq. 共4兲 becomes the same as Sokal’s solution.
Drag can be included by adding a term to the right-hand
side of Eq. 共1兲,
dp
= mg − ␧m␣␷␥ = mg − ␧m␣−␥ p␥ ,
dt
共5兲
Am. J. Phys. 78 共12兲, December 2010
共8兲
ng
,
1 + ␧共1 − n兲/␭
共9兲
where n ⬅ 共1 − ␣兲 / 共2 + ␤ − ␣兲. This acceleration reduces to
Sokal’s drag-free result for ␧ = 0. By including a value for ␧
of the order of 100␭, the acceleration of a raindrop can be
reduced to a more realistic value4 of a few thousandths of g
共for typical values of n of a few tenths兲.
a兲
Electronic mail: mungan@usna.edu
A. D. Sokal, “The falling raindrop, revisited,” Am. J. Phys. 78, 643–645
共2010兲.
2
M. H. Partovi and D. R. Aston, “The generalized raindrop problem,” Am.
J. Phys. 57, 913–920 共1989兲.
3
A similar observation applied to Eq. 共8兲 divided by ␷␤ in Ref. 1 suggests
a change of dependent variable to p = m␷.
4
B. F. Edwards, J. W. Wilder, and E. E. Scime, “Dynamics of falling
raindrops,” Eur. J. Phys. 22, 113–118 共2001兲.
1
with ␧ ⱖ 0 and ␥ ⬎ 0 and where the same exponent ␣ for m is
used for the drag and mass accretion terms because both
effects are expected to scale similarly with the size of
the drop.2 Solving Eqs. 共2兲 and 共5兲 simultaneously is straightforward when ␥ = 1 + ␤, which includes the important special
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g共1 + ␤兲m1−␣
C
+ 共1+␤兲共1+␧/␭兲 .
␭共2 + ␤ − ␣兲 + ␧共1 + ␤兲 m
We let C = 0 共assuming ␷ = 0 when m = 0兲 in Eq. 共8兲, differentiate it with respect to time t, and then substitute Eq. 共2兲 into
the right-hand side and find the acceleration a ⬅ d␷ / dt to be a
constant,
which has the general solution
1+␤
共7兲
as a generalization of Eq. 共3兲. This separated equation can be
integrated. Rewriting the resulting solution in terms of ␷
gives
共2兲
with values of the parameters ␭ ⬎ 0, 1 ⬎ ␣ ⱖ 0, and ␤ ⱖ 0
chosen to ensure stability. Dividing Eq. 共1兲 by Eq. 共2兲 gives
共6兲
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